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Heating effect of electric current

Joule’s law of heating by electric current:

“The heat produced in a conductor is:

(1) Directly proportion the square of current passing through the conductor,(H ∝ I²) keeping R and t constant;

(2) Directly proportion to the resistance of the conductor (H ∝ R) keeping I and t constant;and

(3)Directly proportional to the time of flow of current (H ∝ t) keeping I and R constant.”

Mathematical form of Joule’s law:

   H=I²Rt

Verification of Joule’s law:

The above laws of heating by electric current can be verified in the laboratory using a joule’s calorimeter.

To verify the above laws,heat is generated in a resist;coil R which is enclosed in a copper calorimeter containing water to about two thirds of its volume.The ends of R are connected to finding terminals fixed to the lid.The calorimeter is enclosed in wooden box to minimize the logs of heat.A is the ammeter to measure the current and V is the voltmeter to measure the potential difference across R.A rheostat is connected in the circuit to alter the current.

(a):Let W be the total water equivalent of the calorimeter (including the calorimeter,stirrer and water ) and T1 the initial temperature.A current of I1 amperes is passed for a known interval of time (about 15 to 20 minutes) and the final temperature T2 (after correcting for the loss of heat by radiation ) is noted.Then,the quantity of heat produced =HI=W[θ1 -θ2].Similarly,the experiment is repeated with different values of current I2,I3 etc.( after cooling the calorimeter to the initial temperature) and the corresponding quantities of heat produced H2,H3 etc are noted,keeping R and t constant.

It will be found that:

\frac { { H }_{ 1 } }{ { { I }_{ 1 }^{ 2 } } } =\frac { { H }_{ 2 } }{ { { I }_{ 2 }^{ 2 } } } =\frac { { H }_{ 3 } }{ { { I }_{ 3 }^{ 2 } } } \quad i.e\quad \frac { { H } }{ { { I }^{ 2 } } }

(b):To verify the second law ,the same time t through different resistance coils R1,R2,R3 etc and the corresponding quantities of heat H1,H2,H3 etc are calculated.

Then it will be found that:

\frac { { H }_{ 1 } }{ { { R }_{ 1 } } } =\frac { { H }_{ 2 } }{ { { R }_{ 2 } } } =\frac { { H }_{ 3 } }{ { { R }_{ 3 } } } \quad

(c)To verify the third law,the same current I is passed through the same resistance coil R,for different intervals of time and the quantities of heat produced H1,H2,H3 etc in each case are determined .It will be found that:

\frac { { H }_{ 1 } }{ { { I }_{ 1 } } } =\frac { { H }_{ 2 } }{ { { I }_{ 2 } } } =\frac { { H }_{ 3 } }{ { { I }_{ 3 } } } \quad

What are applications of heating effect of electric current?

1:Safety fuse:A safety fuse is a short piece of wire of low melting point.This is made of an alloy of tin and lead.The fuse wire is connected in series with the electrical installation and,when the current in the circuit exceeds the rated value,the fuse wire gets heated to a temperature higher than its melting point.The wire melts and the electric circuit is cut off.The temperature to which the wire gets heated is directly proportional to the cube of the radius.Thus the temperature depends only on the current flowing through the wire and its radius irrespective of its length.

2:Electric heater or stove,electric Radiators and electric Iron.All these are based on the same principle that is produced when electric current flows through a wire.These contain coils of nichrome (an alloy of nickle and chromium) and large currents of the order of 3 to 5 amperes can be passed through them.In the case of an electric heater working 220 volts and consuming a power of 1000 watts,the current through the coils will be 4.55 amperes.The resistance of the coil will be 48.4 ohms.In the case of electric radiators ,the coil is placed along the axis of a parabolic metallic reflector or at the focus of a concave reflector.

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